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    $\begingroup$ I was so shocked by the assertion that Joel David Hamkins considered (ii) to be better than (i) that I clicked the link to his comment and found that he didn't actually say that. What he says is better than (i) is the $\in$-induction scheme, which is very different from (ii) and in fact implies (i) without any use of choice. For the record, $\in$-induction says that, for every formula $\phi(x)$, possibly with parameters, $\forall x\,[(\forall y\in x\,\phi(x))\to\phi(x)]$ implies $\forall x\,\phi(x)$. $\endgroup$ Commented Nov 11, 2024 at 1:02
  • $\begingroup$ @AndreasBlass: Thanks for pointing that out, I've edited my answer. Apologies to Joel for misrepresenting what he said. $\endgroup$ Commented Nov 11, 2024 at 1:13
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    $\begingroup$ Oops! The "for the record" part of my previous comment contained a typo: $\forall y\in x\,\phi(x)$ should have been $\forall y\in x\,\phi(y)$. $\endgroup$ Commented Nov 11, 2024 at 16:31